Torque, Speed & Power Relation (P = Tω = 2πNT/60)
Calculate mechanical power, motor torque, and rotational speed (RPM) across SI (kW, N·m) and Imperial (HP, lbf·ft) engineering units.
Interactive Calculator Call to Action
Speed Reducer Gearbox Ratio & Torque
Calculate interactively with instant results, step-by-step derivations, and unit conversion. Built on the same engineering standards as this formula reference.
Related Calculators
Primary Mathematical Expression
Nomenclature & Variables
| Symbol | Variable Name | Metric Unit | Imperial Unit | Description |
|---|---|---|---|---|
| P | Mechanical power | kW | hp | The time rate of rotational mechanical work performed. |
| T | Shaft torque | N·m | lbf·ft | The torsional twisting moment transmitted through the rotating element. |
| N | Rotational speed | rpm | rpm | Rotational shaft speed in revolutions per minute (RPM). |
| \omega | Angular velocity | rad/s | rad/s | Rotational angular speed (\omega = 2\pi N / 60). |
Step-by-Step Derivation
- 1
Work done by a constant torque $T$ through an angular rotation angle $\theta$ (in radians) is $W = T \cdot \theta$.
- 2
In one complete revolution ($\theta = 2\pi\text{ rad}$), the mechanical work done is $W_{\text{rev}} = 2\pi \cdot T$.
- 3
For a shaft rotating at $N$ revolutions per minute, the work performed per second (Power $P$ in Watts) is: $P = W_{\text{rev}} \cdot (N / 60) = (2\pi \cdot N \cdot T) / 60 = T \cdot \omega$.
- 4
Converting Power from Watts (W) to kilowatts (kW) where 1 kW = 1,000 W: $P_{\text{kW}} = \frac{2\pi \cdot N \cdot T}{60,000} = \frac{T \cdot N}{9549.296} \approx \frac{T \cdot N}{9550}$.
- 5
In US Customary units where Power is in Horsepower (1 hp = 550 ft·lbf/s) and Torque is in lbf·ft: $P_{\text{hp}} = \frac{2\pi \cdot N \cdot T}{60 \times 550} = \frac{T \cdot N}{5252.113} \approx \frac{T \cdot N}{5252}$.
- 6
When Torque is expressed in inch-pounds (lbf·in): $P_{\text{hp}} = \frac{T_{\text{in·lb}} \cdot N}{63025}$.
Worked Example Calculation
An industrial induction electric motor delivers a rated power output of $15\text{ kW}$ at a synchronous operating speed of $1,450\text{ rpm}$. Calculate the continuous torque delivered by the motor shaft.
- •Identify input parameters: P = 15 kW, N = 1,450 rpm.
- •Select the SI torque relationship: T = (9550 * P) / N.
- •Evaluate torque: T = (9550 * 15) / 1,450 = 143,250 / 1,450 ≈ 98.79 N·m.
- •In Imperial equivalents: 15 kW = 20.11 hp; Torque T = (5252 * 20.11) / 1,450 ≈ 72.85 lbf·ft (874.2 lbf·in).
Engineering Assumptions
- •Steady-state operating conditions with constant rotational speed and uniform torque delivery.
- •Rigid drivetrain connections with zero mechanical slip or clutch disengagement.
- •100% mechanical transmission efficiency (for real gear trains, multiply output power by gear efficiency $\eta_{\text{gear}} \approx 0.95 - 0.98$).
Design Limitations
- •Starting/breakaway torque in electric motors can exceed rated steady-state torque by 200% to 300% (use NEMA / IEC motor service factors).
- •Torsional vibrations and dynamic shock loads require dynamic load multipliers ($K_a, K_o$).
Academic References & Standards
Shigley's Mechanical Engineering Design, 11th Edition
Power transmission fundamentals and rotational mechanics.
Industrial Press
Motor power, speed, torque relationships, and mechanical drive calculations.